Again an electric potential question ?

[Fazza3_uae]

Again an electric potential question ?

Homework Statement

Question # 1

A charge of 7.034 nC is uniformly distributed
along the x-axis from −4 m to 4 m.
What is the electric potential (relative to
zero at infinity) of the point at 5 m on the
x-axis?

Homework Equations

The Attempt at a Solution

I got 21.0728 and its wrong How come ??

 

[willem2]

how did you get this answer?

 

[Fazza3_uae]

The Attempt at a Solution

I know that E in this case = kq/[a*(L+a)]

a : distance from point P to x2 ( P-x2)
L : distance from x1 to x2 ( x2-x1)

The denominator = r²= [a*(L+a)]

I want Potential V=kq/r

r=Sqr(The denominator = r²= [a*(L+a)])

So I just insert the values and got that answer *_* which is wrong

 

[willem2]

I want Potential V=kq/r

V = kq/r is only valid for a point charge or a uniformly charged sphere.
You have to integrate E \bullet ds along a path from the
point with x = 5 to infinity.

 

[Fazza3_uae]

thx 4 clarifying about the potential ^^

But is'nt taking an integral from 5 to infinity will give me infinity and when i multiply infinity with E i will get also infinity ?


In case what u r saying is correct what is the value of r then ??

Is it the distance from point p to x1 or to x2 ? or its something else !

Appreciate Ur Help in Advance Mr. willem2 :)

 

[willem2]

This computation doesn't have an r in it. you already gave an expression for E without
R in it. you can integrate the field along the x axis, so E has the same direction as ds, so you
get

\int_5^{\infty} \frac {k q} {(x-x_1)(x-x_2)}dx


This kind of integral with infinity as a limit is called an improper integral and it is actually a
limit:

\int_a^{\infty} f(x) dx = \lim_{L \to +\infty} \int_a^L f(x) dx

 

[Fazza3_uae]

Aha , i get it now. I will submitt the answer & i am sure it will be correct this time , I'll be right back in minutes ^^

THx again Boss ^^ Really appreciate Ur help !

 

[Fazza3_uae]

Wow nice Job Friend , its absolutly right ^^

thank U very much ^^